Question 13 Marks
If $\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$ and $\frac{\text{d}^2\text{x}}{\text{dt}^2}=\lambda\text{x}$ then find the value of $\lambda.$
Answerwe have $\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=-\text{a}\sin(\text{nt})\times\text{n}-\text{bn}\cos)\text{nt}$$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{an}^2\cos(\text{nt})+\text{bn}^2\sin(\text{nt})$
since, $\frac{\text{d}^2\text{y}}{\text{dt}^2}=\lambda\text{x}$
$\Rightarrow-\text{an}^2\cos(\text{nt})+\text{bn}^2\sin(\text{nt})=\lambda(\text{a}\cos\text{nt}-\text{b}\sin\text{nt})$
$\Rightarrow\lambda=\text{n}^2$
View full question & answer→Question 23 Marks
If $\text{y}=\text{x}+\tan\text{x},$ show that $\cos^2\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{y}+2\text{x}=0$
Answer$\text{y}=\text{x}+\tan\text{x},$
differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\sec^2\text{x}$
differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=0+2\sec^2\times\tan\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\sin\text{x}}{\cos^3\text{x}}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\tan\text{x}+2\text{x}-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2(\text{x}+\tan\text{x})-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{y}-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{y}+2\text{x}=0$
View full question & answer→Question 33 Marks
If $\text{y}=2\sin\text{x}+3\cos\text{x}$ Prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$
Answer$\text{y}=2\sin\text{x}+3\cos\text{x}$Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\cos\text{x}+3(-\sin\text{x})=2\cos\text{x}-3\sin\text{x}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2(-\sin\text{x})-3\cos\text{x}=-(2\sin\text{x}+3\cos\text{x})=\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}} {\text{dx}^2}+\text{y}=0$
View full question & answer→Question 43 Marks
If $\text{y}=\cos^{-1}\text{x},$ Find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ in terms of y alone.
AnswerHere,
$\text{y}=\cos^{-1}\text{x},$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-2\text{x}}{2\sqrt{1-\text{x}^2}^\frac{3}{2}}=\frac{-\text{x}}{(1-\text{x}^2)}$
Now,
$\text{y}=\cos^{-1}\text{x}$
$\Rightarrow\text{x}=\cos\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\cos\text{y}}{(1-\cos^2\text{y})^\frac{3}{2}}=-\frac{\cos\text{y}}{(\sin^2\text{y})^\frac{3}{2}}=-\cot\text{y}\ \text{cosec}^2\text{y}$
View full question & answer→Question 53 Marks
If $\text{x}=\text{a}(\cos2\text{t}+2\text{t}\sin2\text{t})\ \text{and}\ \text{y}=\text{a}(\sin2\text{t}-2\text{t}\cos2\text{t}),$ then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
Answer$\text{x}=\text{a}(\cos2\text{t}+2\text{t}\sin2\text{t})$
$\frac{\text{dx}}{\text{dt}}=-2\text{a}\sin2\text{t}+2\text{a}\sin2\text{t}+4\text{at}\cos2\text{t}=4\text{at}\cos2\text{t}$
$\text{y}=\text{a}(\sin2\text{t}-2\text{t}\cos2\text{t})$
$\frac{\text{dy}}{\text{dt}}=2\text{a}\cos2\text{t}-2\text{a}\cos2\text{t}+4\text{at}\sin2\text{t}=4\text{at}\sin2\text{t}$
$\frac{\text{dy}}{\text{dx}}=\tan2\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}(\tan2\text{t})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sec^22\text{t}\frac{\text{d}}{\text{dx}}(\text{t})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sec^22\text{t}\times\frac{1}{4\text{at}\cos2\text{t}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{2\text{a}}\sec^32\text{t}$
View full question & answer→Question 63 Marks
Find the second order derivatives of the following functions:
$\text{y}=\log(\log\text{x})$
AnswerWe have,
$\text{y}=\log(\log\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{x}}\times\frac{1}{\text{x}}=\frac{1}{\text{x}\log\text{x}}$
Differentiating w.r.t.x, we get
$ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{0-(\log\text{x}+1)}{(\text{x}\log\text{x})^2}=-\frac{(1+\log\text{x})}{(\text{x}\log\text{x})^2}$
View full question & answer→Question 73 Marks
If $\text{y}=\sin(\log\text{x})$ prove that $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
AnswerHere,
$\text{y}=\sin(\log\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\cos(\log\text{x})}{\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\sin(\log\text{x})-\cos(\log\text{x})}{\text{x}^2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\sin(\log\text{x})}{\text{x}^2}-\frac{\cos(\log\text{x})}{\text{x}^2}{}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{y}}{\text{x}^2}-\frac{1}{\text{x}}\times\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
View full question & answer→Question 83 Marks
Find the second order derivatives of the following functions:
$\text{y}=\text{x}^3\log\text{x}$
AnswerWe have
$\text{y}=\text{x}^3\log\text{x}$
differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dy}}=3\text{x}^2\log\text{x}+\text{x}^3\times\frac{1}{\text{x}}$
$=3\text{x}^2\log\text{x}+\text{x}^2$
differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{x}\log\text{x}+3\text{x}^2\times\frac{1}{\text{x}}+2\text{x}$
$=6\text{x}\log\text{x}+5\text{x}$
View full question & answer→Question 93 Marks
If $\text{x}=4\text{z}^2+5,\text{y}=6\text{z}^2+7\text{z}+3$find $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
AnswerHere,
$\text{x}=4\text{z}^2+5,\text{y}=6\text{z}^2+7\text{z}+3$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{dz}}=8\text{z}\ \text{and}\ \frac{\text{dy}}{\text{dz}}=12\text{z}+7$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{12\text{z}+7}{8\text{z}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{12\times8\text{z}-8(12\text{z}+7)}{64\text{z}^2}\times\frac{\text{dz}}{\text{dx}}$
$=\frac{96\text{z}-96\text{z}-56}{512\text{z}^3}=\frac{-7}{64\text{z}^3}$
View full question & answer→Question 103 Marks
If $\text{y}=\log(\sin\text{x})$ Prove that $\frac{\text{d}^3\text{y}}{\text{dx}^3}=2\cos\text{x}\ \text{cosec}^3\text{x}$
AnswerHere,
$\text{y}=\log(\sin\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sin\text{x}}\times\cos\text{x}=\cot\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{cosec}^2\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^3\text{y}}{\text{dx}^3}=-2\text{cosec}\ \text{x}\times(-\text{cosec}\ \text{x}\cot\text{x})$
$=2\cot\ \text{x}\ \text{cosec}^2\text{x}=2\cos\ \text{x}\ \text{cosec}^3\text{x}$
View full question & answer→Question 113 Marks
If $\text{y}=\text{e}^{\text{a}\cos^{-1}}\text{x}$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0$
AnswerHere,
$\text{y}=\text{e}^{\text{a}\cos^{-1}}\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{\text{a}\cos^{-1}}\text{x}\ \times\frac{\text{a}}{\sqrt{1-\text{x}^2}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^{\text{a}\cos^{-1}}\text{x}\ \times\frac{\text{a}^2}{1-\text{x}^2}+\frac{\text{xa }\text{e}^{\text{a}\cos^{-1}}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{y}\times\frac{\text{a}^2}{1-\text{x}^2}-\frac{\text{x}\frac{\text{dy}}{\text{dx}}}{(1-\text{x}^2)}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{a}^2\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0$
View full question & answer→Question 123 Marks
Find the second order derivatives of the following functions:$\sin(\log\text{x})$
AnswerLet $\text{y}=\sin(\log\text{x})$
Then
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}[\sin(\log\text{x})]=\cos(\log\text{x}).\frac{\text{d}}{\text{dx}}(\log\text{x})=\frac{\cos(\log\text{x})}{\text{x}}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big[\frac{\cos(\log\text{x})}{\text{x}}\Big]$
$=\frac{\text{x}.\frac{\text{d}}{\text{dx}}[\cos(\log\text{x})]-\cos(\log\text{x}).\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}$
$=\frac{\text{x}.\Big[-\sin(\log\text{x}).\frac{\text{d}}{\text{dx}(\log\text{x})}\Big]-\cos(\log\text{x}.1}{\text{x}^2}$
$\frac{-\text{x}\sin(\log\text{x}).\frac{1}{\text{x}}-\cos(\log\text{x})}{\text{x}^2}$
$=\frac{[-\sin(\log\text{x})+\cos(\log\text{x})]}{\text{x}^2}$
View full question & answer→Question 133 Marks
If $\log\text{y}=\tan^{-1}$ show that $(1+\text{x}^2)\text{y}_2+(2\text{x}-1)\text{y}_1=0$
AnswerHere$\log\text{y}=\tan^{-1}$
Differentiating w.r.t.x, we get
$\frac{1}{\text{y}}\times\text{y}_1=\frac{1}{1+\text{x}^2}$ $\Rightarrow(1+\text{x}^2)\text{y}_1=\text{y}$ $\Rightarrow(1+\text{x}^2)\text{y}_2+2\text{xy}_1=\text{y}_1$ $\Rightarrow(1+\text{x}^2)\text{y}_2+2\text{xy}_1-\text{y}_1=0$$\Rightarrow(1+\text{x}^2)\text{y}_2+(25\text{x}-1)\text{y}_1=0$
hence proved
View full question & answer→Question 143 Marks
Find the second order derivatives of the following functions:$\text{x}^3+\tan\text{x}$
AnswerWe have
$\text{y}=\text{x}^3+\tan\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{x}}=3\text{x}^2+\sec^2\text{x}$
Differentiating again w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{x}+2\sec^2\text{x}\tan\text{x}$
View full question & answer→Question 153 Marks
If $\text{y}=\frac{\log\text{x}}{\text{x}},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\log\text{x}-3}{\text{x}^3}.$
AnswerHere,
$\text{y}=\frac{\log\text{x}}{\text{x}},$
Differentiating w.r.t.x, we get
$\frac{\text{d}\text{y}}{\text{dx}}=\frac{1-\log\text{x}}{\text{x}^2}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{x}-2\text{x}(1-\log\text{x})}{\text{x}^4}$
$=\frac{-\text{x}-2\text{x}+2\text{x}\log\text{x}}{\text{x}^4}$
$=\frac{-3+2\log\text{x}}{\text{x}^3}$
$=\frac{2\log\text{x}-3}{\text{x}^3}$
Hence proved
View full question & answer→Question 163 Marks
If $\text{y}=\text{ax}^{\text{n+1}}+\text{bx}^{-\text{n}}$ and $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\lambda\text{y}$ then write the value of $\lambda$
Answer$\text{y}=\text{ax}^{\text{n}+1}+\text{b}\text{x}^{-\text{-n}}$
and $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\lambda\text{y}$
Now,
$\frac{\text{dy}}{\text{dx}}=\text{a}(\text{n}+1)\text{x}^{\text{n}}-\text{bn x}^{-\text{n-1}}$
and $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{an}(\text{n}+1)\text{x}^{\text{n}+1}-\text{bn}(-\text{n}-1)\text{x}^{-\text{n}-2}$
Now, $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\lambda\text{y}[\text{given}]$
$\Rightarrow\text{x}^2[\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}]=\lambda(\text{ax}^{\text{n+1}}+\text{b x}^{-\text{n}})$
$\Rightarrow\text{an}(\text{n}+1)\text{x}^{\text{n}+1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}}=\lambda\text{ax}^{\text{n+1}}+\text{b x}^{-\text{n}}$
$\Rightarrow\text{n}(\text{n}+1)\text{ax}^{\text{n}+1}+\text{bx}^{-n}=\lambda\text{ax}^{\text{n}+1}+\text{dx}^{\text{-n}}$
$\Rightarrow\lambda=\text{n}(\text{n}+1)$
View full question & answer→Question 173 Marks
If $\text{y}=\text{e}^\text{x}(\sin\text{x}+\cos\text{x})$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Answer$\text{y}=\text{e}^\text{x}(\sin\text{x}+\cos\text{x})$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})+(\sin\text{x}+\text{cos}\text{x})\text{e}^\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}+\text{e}^\text{x}(\cos\text{x}-\sin\text{x})$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+\text{e}^\text{x}(-\sin\text{x}-\cos\text{x})+(\cos\text{x}-\sin\text{x})\text{e}^\text{x}$
$=\frac{\text{dy}}{\text{dx}}-\text{y}+(\cos\text{x}-\sin\text{x})\text{e}^\text{x}$
Adding and substracting y on RHS
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+\text{e}^\text{x}(-\sin\text{x}-\cos\text{x})+(\cos\text{x}-\sin\text{x})\text{e}^\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
View full question & answer→Question 183 Marks
If $\text{y}=(\tan^{-1}\text{x})^2$ then prove that $(1+\text{x}^2)\text{y}_2+2\text{x}(1+\text{x}^2)\text{y}_1=2$
AnswerHere,
$\text{y}=(\tan^{-1}\text{x})^2$
Differentiating w.r.t.x, we get
$\text{y}_1=\frac{2\tan^{-1}\text{x}}{1+\text{x}^2}$
Differentiating w.r.t.x, we get
$\text{y}_2=\frac{2-4\text{x}\tan^{-1}\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow\text{y}_2=\frac{2}{(1+\text{x}^2)^2}-\frac{2\tan^{-1}\text{x}\times2\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow\text{y}_2=\frac{2}{(1+\text{x}^2)^2}-\frac{2\text{xy}_1}{(1+\text{x}^2)^2}$
$\Rightarrow(1+\text{x}^2)^2\text{y}_2=2-2\text{x}(1+\text{x}^2)\text{y}_1$
$\Rightarrow(1+\text{x}^2)^2\text{y}_2+2\text{x}(1+\text{x}^2)\text{y}_2=2$
Hence proved
View full question & answer→Question 193 Marks
If $\text{y}=\text{e}^{-\text{x}}\cos\text{x},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{e}^{-\text{x}}\sin\text{x}.$
AnswerHere,
$\text{y}=\text{e}^{-\text{x}}\cos\text{x},$
differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{-\text{x}}\cos\text{x}$
$=-\text{e}-\text{x}\sin\text{x}+\text{e}-\text{x}\cos\text{x}$
differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{e}^{-\text{x}}\cos\text{x}-\text{e}^{-\text{x}}\sin\text{x}-\text{e}^{\text{-x}}\cos\text{x}$
$=2\text{e}^{-\text{x}}\sin\text{x}$
View full question & answer→Question 203 Marks
Find the second order derivatives of the following functions:
$\text{y}=\text{x}.\cos\text{x}$
Answerlet $\text{y}=\text{x}.\cos\text{x}$
Then,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}.\cos\text{x})=\cos\text{x}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$=\cos.1+\text{x}(-\sin\text{x})=\cos\text{x}-\text{x}\sin\text{x}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}[\cos\text{x}-\text{x}\sin\text{x}]=\frac{\text{d}}{\text{dx}}(\cos\text{x})-\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{x})$
$=-\sin\text{x}-\Big[\sin\text{x}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$=-\sin\text{x}=(\sin\text{x}+\text{x}\cos\text{x})$
View full question & answer→Question 213 Marks
If $\text{y}=\text{e}^\text{x}\cos\text{x},$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{e}^\text{x}\cos(\text{x}+\frac{\pi}{2}).$
AnswerHere
$\text{y}=\text{e}^\text{x}\cos\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}\cos\text{x}-\text{e}^\text{x}\sin\text{x}=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})+\text{e}^\text{x}(-\sin\text{x}-\cos\text{x})$
$=\text{e}^\text{x}\cos\text{x}-\text{e}^\text{x}\sin\text{x}-\text{e}^\text{x}\sin\text{x}-\text{e}^\text{x}\cos\text{x}$
$=-2\text{e}^\text{x}\sin\text{x}$
$=2\text{e}^\text{x}\cos(\text{x}+\frac{\pi}{2})$
View full question & answer→Question 223 Marks
If $\text{y}=\tan^{-1}$ show that $(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{x}\frac{\text{dy}}{\text{dx}}=0$
Answer$\text{y}=\tan^{-1}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
$\Rightarrow(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}=1$
Differentiating w.r.t.x, we get
$\Rightarrow(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{x}\frac{\text{dy}}{\text{dx}}=0$
Hence proved
View full question & answer→Question 233 Marks
If $\text{y}=3\text{e}^{2\text{x}}+2\text{e}^{3\text{x}}$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}-5\frac{\text{dy}}{\text{dx}}+6\text{y}=0$
Answer$\text{y}=3\text{e}^{2\text{x}}+2\text{e}^{3\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=6\text{e}^{2\text{x}}+6\text{e}^{3\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=12\text{e}^{2\text{x}}+18\text{e}^{3\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=5(6\text{e}^{2\text{x}}+6\text{e}^{3\text{x}})-6(3\text{e}^{2\text{x}}+2\text{e}^{3\text{x}})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=5\Big(\frac{\text{dy}}{\text{dx}}\Big)-6\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-5\Big(\frac{\text{dy}}{\text{dx}}\Big)+6\text{y}=0$
View full question & answer→Question 243 Marks
If $\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
AnswerHere,
$\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=2\text{a}\text{e}^{2\text{x}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\text{a}\text{e}^{2\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{a}\text{e}^{2\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
Hence proved.
View full question & answer→